UVa 10539 (筛素数、二分查找) Almost Prime Numbers

题意：

求正整数L和U之间有多少个整数x满足形如x=p^k 这种形式，其中p为素数，k＞1

分析：

首先筛出1e6内的素数，枚举每个素数求出1e12内所有满足条件的数，然后排序。

对于L和U，二分查找出小于U和L的最大数的下标，作差即可得到答案。

#include <cstdio>
#include <cmath>
#include <algorithm>

typedef long long LL;

const int maxn = 1000000;
const int maxp = 80000;

bool vis[maxn + 10];
int prime[maxp], cntp = 0;
LL a[maxn], cnt = 1;

void Init()
{
    int m = 1000;
    for(int i = 2; i <= m; ++i) if(!vis[i])
        for(int j = i * i; j <= maxn; j += i) vis[j] = true;
    for(int i = 2; i <= maxn; ++i) if(!vis[i]) prime[cntp++] = i;

    for(int i = 0; i < cntp; ++i)
    {
        LL temp = (LL)prime[i] * prime[i];
        while(temp <= 1000000000000LL)
        {
            a[cnt++] = temp;
            temp *= prime[i];
        }
    }
    std::sort(a, a + cnt);
}

int binary_search(LL n)
{
    LL L = 0, R = cnt;
    while(L < R)
    {
        LL mid = ((L + R + 1) >> 1);
        if(a[mid] <= n) L = mid;
        else R = mid - 1;
    }
    return L;
}

int main()
{
    Init();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        LL hehe, haha;
        scanf("%lld%lld", &hehe, &haha);
        int t = binary_search(hehe), k = binary_search(haha);
        int ans = k - t;
        if(a[t] == hehe) ans++;
        printf("%d\n", ans);
    }

    return 0;
}